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04 Oct 2016, 06:52
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If x and y are positive integers, when x^4-y^4 is divided by 4, what is the remainder?
1) x-y is divisible by 4
2) x^2+y^2 is divisible by 4
==> If you modify the original condition and the problem, you get x^4-y^4=(x-y)(x+y)(x^2+y^2). Then, from 1)=2), the remainder of what are both divided by 4 is 0, hence unique, and sufficient. The answer is D.
Answer: D
1) x-y is divisible by 4
2) x^2+y^2 is divisible by 4
==> If you modify the original condition and the problem, you get x^4-y^4=(x-y)(x+y)(x^2+y^2). Then, from 1)=2), the remainder of what are both divided by 4 is 0, hence unique, and sufficient. The answer is D.
Answer: D
11 Oct 2016, 04:50
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The question below is also a 5051-level question, and an inequality problem that ignores what is squared.
Is 1/x > 1/x^2?
1) x>0
2) 1>1/x
==> If you modify the original condition and the problem, and square the inequality equation, what is squared is always a positive number. Hence, it does not change the sign of inequality. So, if you multiply both sides by x^2, you get x>1?. Since there is 1 variable, D is highly likely to be the answer. Since the range of the question should include the range of condition in order for the condition to be sufficient, in the case of 1), the range of the question does not include the range of the condition, hence not sufficient. In the case of 2), if you multiply both sides of the equation by x^2, you get x2>x, and from x(x-1)>0, then x<0 or 1<x. From this, the range of the question does not include the range of the condition, hence not sufficient. In the case of 1) & 2), from x>1, the range of the question includes the range of the condition, hence sufficient. The answer is C.
Answer: C
Is 1/x > 1/x^2?
1) x>0
2) 1>1/x
==> If you modify the original condition and the problem, and square the inequality equation, what is squared is always a positive number. Hence, it does not change the sign of inequality. So, if you multiply both sides by x^2, you get x>1?. Since there is 1 variable, D is highly likely to be the answer. Since the range of the question should include the range of condition in order for the condition to be sufficient, in the case of 1), the range of the question does not include the range of the condition, hence not sufficient. In the case of 2), if you multiply both sides of the equation by x^2, you get x2>x, and from x(x-1)>0, then x<0 or 1<x. From this, the range of the question does not include the range of the condition, hence not sufficient. In the case of 1) & 2), from x>1, the range of the question includes the range of the condition, hence sufficient. The answer is C.
Answer: C
17 Oct 2016, 00:26
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Also, you must be wary of CMT 4(A) question below.
(integer) If a and b are integers, is anodd number?
1) a+b is an even number
2) ab is an odd number
==>In the original condition, there are 2 variables, so C is the answer. Through 1) & 2), a=b=odd, so yes, hence sufficient. The answer is C. This problem, too, is an integer problem, which is a key question, so if you apply CMT 4(A), and tackle 2), a=b=odd, hence yes, and sufficient. B is the answer.
Answer: B
(integer) If a and b are integers, is anodd number?
1) a+b is an even number
2) ab is an odd number
==>In the original condition, there are 2 variables, so C is the answer. Through 1) & 2), a=b=odd, so yes, hence sufficient. The answer is C. This problem, too, is an integer problem, which is a key question, so if you apply CMT 4(A), and tackle 2), a=b=odd, hence yes, and sufficient. B is the answer.
Answer: B
